Longest path on a 2d plane

Question

You are provided a set of arbitary, unique, 2d, integer Cartesian coordinates: e.g. [(0,0), (0,1), (1,0)]

Find the longest path possible from this set of coordinates, with the restriction that a coordinate can be "visited" only once. (And you don't "come back" to the coordinate you started at).

Important:

You cannot "pass over" a coordinate or around it. For instance, in the last note example (Rectangle), you cannot move from D to A without visiting C (which may be a revisit, invalidating the length thus found). This was pointed out by @FryAmTheEggman.

Function Input: Array of 2d Cartesian Coordinates
Function Output: Maximum length only
Winner: Shortest code wins, no holds barred (Not the most space-time efficient)

---

Examples

1: In this case shown above, the longest path with no coordinate "visited" twice is A -> B -> O (or O-B-A, or B-A-O), and the path length is sqrt(2) + 1 = 2.414

---

2: In this case shown above, the longest path with no coordinate "visited" twice is A-B-O-C (and obviously C-O-B-A, O-C-A-B etc.), and for the unit square shown, it calculates to sqrt(2) + sqrt(2) + 1 = 3.828.

---

Note: Here's an additional test case which isn't as trivial as the two previous examples. This is a rectangle formed from 6 coordinates:

Here, the longest path is: A -> E -> C -> O -> D -> B, which is 8.7147
(max possible diagonals walked and no edges traversed)

Answer 1

Pyth, 105 103 100 92 86 bytes

V.pQK0FktlNJ.a[@Nk@Nhk)FdlNI&!qdk&!qdhkq+.a[@Nk@Nd).a[@Nd@Nhk)J=K.n5B)=K+KJ)IgKZ=ZK))Z

              Z = 0 - value of longest path
              Q = eval(input())

V.pQ         for N in permutations(Q):
  K0           K = 0 - value of current path
  FktlN        for k in len(N) - 1:
    J.a          set J = distance of
    [@Nk                 Q[k] and Q[k+1]
    @Nhk)    
    FdlN         for d in len(N):
I&                 if d != k && d != (k + 1)
!qdk
&!qdhk
q+                and if sum of
.a                   distance Q[k] and Q[d]
 [@Nk                
 @Nd)                
.a                   distance Q[d] and Q[k+1]
 [@Nd
 @Nhk)
J                    are equal to J then
  =K.n5              set K to -Infinity
  B                  and break loop
                     ( it means that we passed over point )
  )                   end of two if statements
=K+KJ                  K+=J add distance to our length
)                      end of for
IgKZ                   if K >= Z - if we found same or better path
  =ZK                  Z = K       set it to out max variable
))                     end of two for statements
Z                      output value of longest path 

Try it here!

Answer 2

Mathematica, 139 bytes

Max[Tr@BlockMap[If[1##&@@(Im[#/#2]&@@@Outer[#/Abs@#&[#-#2]&,l~Complement~#,#])==0,-∞,Abs[{1,-1}.#]]&,#,2,1]&/@Permutations[l=#+I#2&@@@#]]&

---

Test case

%[{{0,0},{0,1},{1,0},{1,1},{2,0},{2,1}}]
(* 3 Sqrt[2]+2 Sqrt[5] *)

%//N
(* 8.71478 *)

Answer 3

Perl, 341 322 318 bytes

sub f{@g=map{$_<10?"0$_":$_}0..$#_;$"=',';@l=grep{"@g"eq join$",sort/../g}glob"{@g}"x(@i=@_);map{@c=/../g;$s=0;$v=1;for$k(1..$#c){$s+=$D=d($k-1,$k);$_!=$k&&$_!=$k-1&&$D==d($_,$k)+d($_,$k-1)and$v=0 for 0..$#c}$m=$s if$m<$s&&$v}@l;$m}sub d{@a=@{$i[$c[$_[0]]]};@b=@{$i[$c[$_[1]]]};sqrt(($a[0]-$b[0])**2+($a[1]-$b[1])**2)}

The code supports up to a 100 points. Since it produces all possible point permutations, 100 points would require at least 3.7×10¹³⁴ yottabytes of memory (12 points would use 1.8Gb).

Commented:

sub f {
    @g = map { $_<10 ? "0$_" : $_ } 0..$#_; # generate fixed-width path indices
    $" = ',';                               # set $LIST_SEPARATOR to comma for glob
    @l = grep {                             # only iterate paths with unique points
        "@g" eq join $", sort /../g         # compare sorted indices with unique indices
    } glob "{@g}" x (@i=@_);                # produce all permutations of path indices
                                            # and save @_ in @i for sub d
    map {
        @c = /../g;                         # unpack the path indices
        $s=0;                               # total path length
        $v=1;                               # validity flag
        for $k (1..$#c) {                   # iterate path
            $s +=                           # sum path length
                $D = d( $k-1, $k );         # line distance 

              $_!=$k && $_!=$k-1            # except for the current line,
              && $D == d( $_, $k )          # if the point is on the line,
                     + d( $_, $k-1 )
              and $v = 0                    # then reset it's validity
            for 0 .. $#c                    # iterate path again to check all points
        }
        $m=$s if $m<$s && $v                # update maximum path length
    } @l;
    $m                                      # return the max
}

sub d {                                     
    @a = @{ $i[$c[$_[0]]] };                # resolve the index $_[0] to the first coord
    @b = @{ $i[$c[$_[1]]] };                # idem for $_[1]
    sqrt( ($a[0] - $b[0])**2       
        + ($a[1] - $b[1])**2 )      
}

TestCases:

print f( [0,1], [0,0], [1,0] ), $/;        $m=0; # reset max for next call
print f( [0,0], [0,1], [1,0], [1,1] ), $/; $m=0;
print f( [0,0], [0,1], [0,2] ), $/;        $m=0;
print f( [0,0], [0,1], [0,2], 
         [1,0], [1,1], [1,2]),$/;          $m=0;

• 322 bytes: save 19 by not resetting $", and some inlining
• 318 bytes: save 4 by reducing max nr of coords to 100.